Proving $(a^n + b^n) \sum_{k=0}^{n} \binom{n}{k} a^k b^{n-k} \leq (a^{2n} + b^{2n}) \sum_{k=0}^{n} \binom{n}{k}$

Question

I'm currently preparing for a French math contest, and I'm having some trouble proving an inequality. The inequality I'm trying to show is:

Problem. Show that \begin{equation} (a^n + b^n) \sum_{k=0}^{n} \binom{n}{k} \cdot a^k \cdot b^{n-k} \leq (a^{2n} + b^{2n}) \sum_{k=0}^{n} \binom{n}{k} \end{equation} Here, both $a$ and $b$ are assumed to be positive real numbers.

We can write this as:

\begin{equation} (a^n + b^n) (a+b)^{n} \leq 2^{n}(a^{2n} + b^{2n}) \end{equation}

I believe I need to prove that:

\begin{equation} a^{k+n} b^{n-k} + a^k b^{2n-k} \leq a^{2n} + b^{2n} \end{equation}

I would really appreciate it if someone could help me with this proof. Any insights or hints on how to approach this problem would be greatly appreciated. Thank you in advance for your assistance.

Best regards.

Answer 1

Your guess is almost correct. By noting that

\begin{align*} (a^n + b^n) \sum_{k=0}^{n} \binom{n}{k} a^k b^{n-k} &= a^n \sum_{k=0}^{n} \binom{n}{k} a^k b^{n-k} + b^n \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \\ &= \sum_{k=0}^{n} \binom{n}{k} (a^{n+k}b^{n-k} + a^{n-k}b^{n+k}), \end{align*}

it suffices to prove that

$$ a^{n+k}b^{n-k} + a^{n-k}b^{n+k} \leq a^{2n} + b^{2n} \tag{*} $$

for each $k = 0, \ldots, n$. To this end, note that for any non-negative integers $p$ and $q$,

$$ a^p b^q = \bigl[ \underbrace{a^{p+q} \cdot a^{p+q} \cdots a^{p+q}}_{p\text{ times}}\cdot\underbrace{b^{p+q} \cdot b^{p+q} \cdots b^{p+q}}_{q\text{ times}} \bigr]^{\frac{1}{p+q}} \leq \frac{p a^{p+q} + q b^{p+q}}{p+q} $$

by the AM-GM inequality.¹⁾ By applying this inequality, we get

\begin{align*} a^{n+k}b^{n-k} + a^{n-k}b^{n+k} &\leq \frac{(n+k)a^{2n} + (n-k)b^{2n}}{2n} + \frac{(n-k)a^{2n} + (n+k)b^{2n}}{2n} \\ &= a^{2n} + b^{2n} \end{align*}

as required.

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^{1) This inequality actually holds for any real $p, q \geq 0$ so that $p +q > 0$, which can be proved using Young's inequality or Jensen's inequality. However, we don't need this for our proof.}

Answer 2

A less tricky approach is (letting $x:=\frac a{a+b}$) to prove that $$\forall x\in(0,1),\quad f(x):=2^n\left(x^{2n}+(1-x)^{2n}\right)-x^n-(1-x)^n\ge0.$$ For this, just notice that $f(1/2)=0$ and $f'$ is negative on $(0,1/2)$ and positive on $(1/2,1).$

Answer 3

We have $$a^{2n} + b^{2n} - a^{n + k}b^{n-k} - a^{n-k}b^{n+k} = (a^{n + k} - b^{n + k})(a^{n - k} - b^{n - k})\ge 0.$$ (Note: The last inequality is easy. WLOG, assume that $a \ge b$. Then $a^{n+k} \ge b^{n+k}$ and $a^{n-k} \ge b^{n-k}$. We are done. By the way, I used this trick several times, e.g. here.)